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User talk:Mush9
Ohai. Leave a message if you want. Mush9 (talk) 19:48, January 21, 2017 (UTC) New user, new googologlists. AarexWikia04 - 17:41, October 7, 2016 (UTC) And it's allways exaiting!Boboris02 (talk) 16:46, October 12, 2016 (UTC) I can help you with NOOP notation... My idea is that you can construct large self-accelarating set structures in ZFC as your starting point.We can even make your own type of theory similar to ZFC.I'm going to refer to it as MOT (Mush9's Output Theory). Let' \(z\) in MOT be the smallest nonnegative finite integer,that is not accesable with just the logical connectivities used in ZFC in \(n\) characters or less.(these are \(\land,\lor,\neg,\Rightarrow,\Leftrightarrow,\forall,\exists,(,)\)) will be the first unprovable set in a \(a \land^{n-1} b\) languige. Right now this is ill-defined,because "a \(a \land^{n-1}\) languige" is not properly defined. But we can do so! Let \(\forall b \exists z > \Delta_xb \) be the Axiom of functions,which will be our first statement for MOT. This means for any integer \(b\),there is an integer \(z\) which is greater than any function \(\Delta_x\) which is describable in no less than \(b\) symbols (describing any symbol as a single bit) in an \(\Phi(n)=\Phi_1(0) \neg n\) -system and \(x\) is an integer povable in ZFC in finite steps. This theory is quite new,but perhaps we could extend it furthur together,if you like my idea! Are you up to the challenge? Boboris02 (talk) 16:33, November 12, 2016 (UTC) In my previous message I talked about MOT and more spesificaly \(a \land^n b\) languige. Now,I'm going to define that! \(a \land^1 b = a \land b\),if we have two sets \(S_1\) and \(S_2\) and the intersection of the two sets is equal to a,while the parts excluded equal b,then: \(\forall a \in S_1,a \in S_2,b =S_1 \notin S_2 : a \land b > \underbrace{\{\ldots,a,a,a,\ldots\}}_{b-1} \in y\) That will be our Axiom of length. \(\Delta_xb\) System Next,let's define something that does not exist in ZFC,but it exists in MOT: Using the Axiom of functions,that I described previously \(\forall b \exists z > \Delta_xb\) , we can go more in detail about what \(\Delta_xb\) realy is.In MOT \(\Delta_n\) allways accounts for a function,infact \(\Delta\) in general will be the symbol used to mark,that the type of operation is function. By the way in MOT there are 6 types of operations:Statments,Theorems,Calculations,Functions,Systems and Notators.But we will get to that later... Back to where I was:\(\Delta_xb\) means a function,that we named \(\Delta_x\) which is describable in \(b\) many symbols in a \(\Phi(n)=\Phi_1(0) \neg n\) -system(accounting every number as a single symbol and not counting \(()\) brackets as symbols) Now what is a \(\Phi(n)=\Phi_1(0) \neg n\) -system? Phi(\(\Phi(n)\)) Systems In MOT there are many phi systems. Phi systems are ways to show how a function works and what type of function it is. Example:What would the Ackerman function(\(Ack(a,b)\)) be? It would be a \(\Phi(a)\Rightarrow b=\Phi(a-1)\Rightarrow (\Phi(a)\Rightarrow (b-1))\) system. What would four-entry arrays in BEAF(\(\{a,b,c,d\}\)) be? It would be a... \(\Phi(a) \Rightarrow b \Rightarrow c \Rightarrow d = \Phi(a) \Rightarrow (\Phi(a) \Rightarrow (b-1) \Rightarrow c \Rightarrow d) \Rightarrow (c-1) \Rightarrow d\) -system If at some point the system changes,it will have to be pointed out that the system has changed and the primary system will be sent back inside the secondary system. Example:What would \(f_3(n)\) be? It would be \(\Phi_3(n)\) inside a \(\Phi_x(n) = \Phi^{n}_{(x-1)}(n)\) -system inside a \(\Phi_0(n) = n+1\) -system Pretty easy....right? Well at this level,yes!... But soon we will come into things,where there is no obvious output! But for now lets concentrate on easier ones,but add something new! If we add the \(\cup \) operator into this,the results come scary:For instance,a \(\Phi(n) = \Phi(n-1) \cup 1\) -system has a growth rate of \(f_{\omega}(n)\) and a \(\Phi(n) = \Phi(n-1) \cup n\) -system has a growth rate of \(\approx f_{\omega^{\omega}}(n)\). Now I will explain the operators: \(\Rightarrow n\) means repeating everything previously set in the operation \(n\) times. \(\Rightarrow n\Rightarrow m\) means repeating the system n,m times. \(\Rightarrow n\Rightarrow m\Rightarrow k\) means repeating the system \(n\Rightarrow m\), k times. \(a \cup 1\) means a long chain of repetitions \(\Phi(a) \cup 1 = \Phi(a) \Rightarrow (\Phi(a)\Rightarrow a \Rightarrow \ldots \Rightarrow a)\Rightarrow a\) \(\Phi(a) \cup 2 = \Phi(a) \cup 1 \Rightarrow \Phi(a-1) \cup 2) \Rightarrow a\) With just these operators we can set any system we want: \(\Phi(a) \Rightarrow b = \Phi^b(a) \Rightarrow (\Phi(a) \Rightarrow (b-1)) \cup a \Rightarrow a\) -system for example is a system whith a growth rate of something like \(f_{\psi(\Omega^{\Omega^{\varepsilon_0}+b})}(a)\) Another good system is \(\Phi(a) \Rightarrow b \cup c \Rightarrow d = \Phi^b(a) \Rightarrow (\Phi(a) \cup b) \Rightarrow c \cup \Phi_c(b-1)\) where \(\Phi_a(b) = \Phi_a(b-1) \Rightarrow a \cup b\) This has a growth rate of \(f_{\psi(\psi_I(0))}(a)\) The problem is that we cannot make uncomputable functions like this.....right? Well,if I introduse you to the two new operators \(\neg n\) and \(\Leftrightarrow n\) it is! \(\Phi(n) \neg b\) on it's own does not mean anything,but it can be placed in some systems and there it's meaning can be created Eg.\(\Phi(n) \neg a = \Phi(n) >(+1) \Phi(n) \neg (a-1)\) which means it overgrows the other system by one on each go! That way we can set it defalt as eventualy overgrowing a system like so: \(\Phi(n) \neg a >^*(+\Phi(n-1)) \Phi(n) \neg (a-1) \) This way it will eventualy overgrow the other system by the previous variant. \(\Phi(n) \neg a \Leftrightarrow b\) means that a turing machine will not halt untill the b-th turn. Lets see how we can create an uncomputable function.... \(\Phi(n) = n \neg n \Leftrightarrow k \Leftrightarrow m = {\ldots 000 \underbrace{1111\ldots 111}_m 000\ldots} \Leftrightarrow 0 = \infty\) What you're seeing is the system that represents \(\Sigma(n)\),the busy beaver function,where the tape is infinite and is completely set to zeros. I'll continue next time.I was just making things clear. Boboris02 (talk) 17:22, November 13, 2016 (UTC) Brilliant! I have a little look. Mush9 (talk) 18:56, November 13, 2016 (UTC) Thanks! I'm happy you liked it! I will add my other ideas later.Boboris02 (talk) 21:13, November 13, 2016 (UTC) Okay we got pretty far last time,but I want to work more on the \(\Phi(n)\) systems. Last time I defined the busy beaver function with a \(\Phi(n)\) -system,now I want to show you a way of using ordinals inside phi systems. \(\Phi(n) \neg n = \Phi(n) \Leftrightarrow | \mathbb{U} |\) ,which means this function is uncomputable,denoted with the \(| \mathbb{U} |\) symbol. \(| \mathbb{U}^{>^*} \mathbb{C} |\) means it's an uncomputable function,which overgrows and eventualy dominates all computable functions.The symbol \(| \mathbb{C} |\) means computable. \(\Phi(n) \neg n = \Phi(n) \Leftrightarrow | \mathbb{U}^{>^*} \lor \mathbb{N} | \) means the minimum growth rate needed(as shown by \(\lor\),the minimal symbol) to eventually overtake another function and \(| \mathbb{N} |\) is an arbitrary function. Boboris02 (talk) 16:59, November 14, 2016 (UTC) I like your idea of phi systems. Nishada 23:19, November 14, 2016 (UTC) : I agreed. AarexWikia04 - 23:28, November 14, 2016 (UTC) Ordinal functions inside Phi systems Well now we can put systems for spessific growth rates! \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \mathbb{C}|\) is the smallest possible uncomputable function,that eventualy dominates all computable functions in the languige of the phi systems,thus it has a growth rate of \(f_{\omega^{\text{CK}}_1}(n)\),the Church-Kleene ordinal. \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \Phi(n-1) |\),then would be \(f_{\omega^{\text{CK}}_{\omega}}(n)\),which would be the nth function that dominates the previous one,but how do we get more powerfull than that? A \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \Phi(\Phi(n)) |\) system would loop forever,without giving an answer and as it turns out there is no way of making a phi system that grows faster than \(\omega^{\text{CK}}_{\omega}\) ,without using subscripts! Subscripts are a way to diferantiate systems,that collapse on themselves! A \(\Phi_1(n) = n \neg n | \mathbb{U}^{>^*} \Phi(\Phi_1(n-1)) |\) -system inside of an \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \mathbb{C} |\) -system dominates all recursive non-recursive ordinals and it has an ordinal that is non-recursively non-recursive! It reaches \(\omega^{\text{CK}}_{\omega^{\text{CK}}_1}\) And if we change the primary system from \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \mathbb{C} |\) -system to a \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \Phi(n-1) |\) -system,it goes up to \(\omega^{\text{CK}}_{\omega^{\text{CK}}_{\omega}}\) If we continue with higher subscripts,like: \(\Phi_m(n) = n \neg n | \mathbb{U}^{>^*} \Phi_{m-1}(\Phi_k(n-1)) |\) ,we can climb into big nestings of Church-Kleene-like structures! To express the value of the above \(\Phi_m(n)\) we have to come up with a new ordinal-collapsing functions! Let \(\Lambda_0 = \omega^{\text{CK}}_1\) and \(\Lambda_k = \omega^{\text{CK}}_{\Lambda_{k-1}}\) ,then \(\Phi_k(n) = f_{\Lambda_k}(n)\) and \(\Phi_n(n) = f_{\Lambda_{\omega}}(n)\). Finally,we come to the conclusion,that the max growth rate of the phi systems is \(\Lambda_{\omega}(n)\)......WRONG! \(\Phi_m(n) = n \neg n | \mathbb{U}^{>^*} \Phi_{m-1}(\Phi_k(n-1)) |\) is not the most powerful system! \(\Phi(n) = \Phi_1(0) \neg n\) is WAY more powerful than \(\Lambda_{\omega}(n)\)!So much more,it's hardly even descibable! But we can even extend the system more with adding larger subscripts! \(\Phi_m(n) = \Phi_{m+1}(0) \neg n\). Trying to work out the growth rate of those is an unbelievable task,but i will try just for the sake of it! \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \mathbb{C}|\) growth rate of \(\omega^{\text{CK}}_1\) \(\Phi_1(0) \neg n = \Phi(n)\) growth rate of \(\omega^{\text{CK}}_1\) \(\Phi_1(0) \neg n \neg n = \Phi(n) \neg n\) growth rate of \(\omega^{\text{CK}}_{\omega^{\text{CK}}_1}\) or \(\Lambda_1\) \(\Phi_1(0) \underbrace{\neg n \neg n \ldots \neg n}_n\) has a growth rate of \(\Lambda_{\omega}\) \(\Phi_1(0) = \Phi_1(0) \neg 0\),since one of the rules of negation in MOT is that \(a \neg b \neg \ldots \neg n \neg 0 = a \neg b \neg \ldots \neg n\),the zero at the end can be removed. Thus \(\Phi_1(0) = \Phi_1(0) \neg 0 = \Phi_2(0) = \ldots = \Phi_n(0) \) and all \(\Phi_n(0) = \Phi(0) = 0\),no growth rate! Using that statement and the reverse of all other previous statemants,then: \(\Phi_1(1) = \Phi_1(1) \neg 0 = \Phi_2(0) \neg 1 \neg 0 \ldots\) \( = \Phi_n(0) \underbrace{\neg 0 \neg 0 \ldots \neg 0}_{n-2} \neg 1 \neg 0\) and \(\Phi_1(1) = \Phi_1(1) \neg 0 \neg \ldots = \Phi(\Phi(\Phi(\ldots(\Phi(1) \neg 1) \neg 1)\ldots )\neg 1\) Finaly,the growth rate of that is an infinite nesting of my Lambda function and thus \(\Phi_1(1)\) marks the supremum of the lambda function! This realy makes you think about the sheer size of what we're dealing with here... We thought \(\Phi(n) = n \neg n | \mathbb{U}^{>^*} \lor \Phi(n-1) |\) was the best we could do with a growth rate of \(\omega^{\text{CK}}_{\omega}\),then it was topped by \(\Phi_1(n) = n \neg n | \mathbb{U}^{>^*} \lor \mathbb{C} |\) with a growth rate of \(\omega^{\text{CK}}_{\omega^{\text{CK}}_1}\) or \(\Lambda_1\) and later by \(\Phi_1(n) = n \neg n | \mathbb{U}^{>^*} \lor \Phi(n-1) |\) with a growth rate of \(\omega^{\text{CK}}_{\omega^{\text{CK}}_{\omega}}\) and then by \(\Phi_m(n) = n \neg n | \mathbb{U}^{>^*} \Phi_{m-1}(\Phi_k(n-1)) |\),with a growth rate of \(\Lambda_{\omega}\) But they are all easily destroyed by \(\Phi_1(1)\)! What exactly is the growth rate of \(\Phi_1(1)\)? Well.....it marks the supremum of my notation and it is the limit of \(\{\Lambda_0,\Lambda_{\Lambda_0},\Lambda_{\Lambda_{\Lambda_0},\ldots}\}\)!!! And lastly,remember \(\Phi_1(1)\) is just the very first and tiniest of the \(\Phi_1(n)\) sequense ,which on it's own is just the first and tiniest of the \(\Phi_n(k)\) sequence..... Boboris02 (talk) 21:33, November 15, 2016 (UTC) If you're confident - publish it. I'm currently creating the KING function. Mush9 (talk) 08:51, November 25, 2016 (UTC) ....So are you abandoning MOT? Boboris02 (talk) 14:29, December 2, 2016 (UTC) It's your idea - you know way better than me about it :P Mush9 (talk) 18:14, December 2, 2016 (UTC) But......but.....I didn't publish it because you didn't add anything but it was your idea so it would come out as if I was stealing your idea.So if I publish it I will say that NOOP languige,notation and function and the thing that inspired me to do this was your idea and if you didn't write that blog post I would've never created MOT.....or should I refer to it as MBOT?Anyways,even if you haven't added anything,you've still contributed A LOT because without you it wouldn't exist.So if I ever publish it it will have a reference that you began the whole thing and made it happen! Boboris02 (talk) 15:50, December 11, 2016 (UTC) Well... thanks. I think you should publish it, with or without the reference (which I feel isn't needed). However, do whatever you want. Over Christmas, I'll have a little read and check it and all that :P Mush9 (talk) 17:34, December 12, 2016 (UTC) KING I'd like to make a general comment about the idea of your KING function, or more precisely about orbles. Just saying that they are "objects such that blah blah", you are imposing nearly no structure on them. To explain what I mean, contrast this with the notion of a set. The "collection of all sets" cannot be just an arbitrary collection of things with a relation (\(\in\)) between them, but it has to have certain properties, namely it has to satisfy . Without an analogue of these for orbles, we could just postulate that a collection containing just one "orble" is the collection of all orbles. Now you might think to yourself "but you did something like that with oodles, so why can you do something like that and I can't!?". The answer is, I didn't, as I explain here. Some things said there can also be applied here - when you have thought of the orbles, you might have had an impression that somewhere, "out there", in some "ambient space", the orbles exist just as you have dreamt of them. This is not necessarily the case. For FOOT, I have defined certain objects which are all parts of the universe of sets, and I'd recommend for you to do something similar - to describe orbles in terms of sets. Good luck with your project. LittlePeng9 (talk) 21:24, November 28, 2016 (UTC) Yep. I feel that every function that has attempted to overtake FOOT has in one way or another copied it! Still, I'll try to make this work. I'll rethink it and bear everything you've said in mind. Mush9 (talk) 23:05, November 28, 2016 (UTC) Upon your request I have deleted that blog post. If you weren't being serious about it, I can still undelete it. LittlePeng9 (talk) 20:35, December 3, 2016 (UTC) Thanks, Peng. I'm going to use say, Google Sites, until I'm sure the function is worthy of publishing. Mush9 (talk) 10:00, December 4, 2016 (UTC) I've got a website, now: https://mush9.github.io/googology/ Mush9 (talk) 17:31, December 4, 2016 (UTC) NOOP function is finaly defined! I wrote a blog post about it.Now MBOT is fully finished!Boboris02 (talk) 19:14, December 17, 2016 (UTC) Hooray! Mush9 (talk) 10:04, December 18, 2016 (UTC) Not gonna lie - that kind of failed :P It's okay though. If you want you can help me out with king - see profile page Mush9 (talk) 10:31, December 19, 2016 (UTC) IRC Yes, \(a=b\Leftrightarrow a=c\) is a statement, or, more precisely, a formula (afaik the term "statement" is not well-established in logic), with three free variables \(a,b,c\). Also, you know what I really dislike? When people suddenly leave the chat without any prior warning. They don't even leave any time to say bye to them :( LittlePeng9 (talk) 20:57, January 21, 2017 (UTC) Sorry, I'll try not leave spontaneously again. What I was trying to say, before I rudely left, was "what does Emlightened mean by \(a = \langle b,c,d,e\rangle \leftrightarrow a=\langle \langle b,c,d \rangle ,e \rangle \) and \(On(a) \leftrightarrow \forall b\in a(\cup b\in a \wedge\forall c\in b(\cup c \in b))\)"? Mush9 (talk) 09:14, January 22, 2017 (UTC) :These are actually definitions - she defines what the statements \(a = \langle b,c,d,e\rangle\) and \(On(a)\) mean. This is done for convenience and human readability, but it isn't strictly necessary - we could've replaced each instance of \(On(x)\) by the formula \(\forall b\in x\dots\) and similarly for ordered tuples, but we don't so that the axiom is at least somewhat readable. LittlePeng9 (talk) 10:50, January 22, 2017 (UTC)